Question: The equation of an ellipse is given below. $\dfrac{(x+7)^2}{8}+\dfrac{(y-10)^2}{13}=1$ What are the foci of this ellipse ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-7,10+\sqrt{5})$ and $(-7,10-\sqrt{5})$ (Choice B) B $(7+\sqrt{105},-10)$ and $(7-\sqrt{105},-10)$ (Choice C) C $(-7+\sqrt{5},10)$ and $(-7-\sqrt{5},10)$ (Choice D) D $(7,-10+\sqrt{105})$ and $(7,-10-\sqrt{105})$
Explanation: The strategy The foci of an ellipse are always located on its major axis, and are the same distance away from the center of the ellipse. In order to find the foci, we need to know the distance between a focus and the center, which is also known as the focal length. We can find this distance by using the formula $f^2=p^2-q^2$, where $f$ represents the distance between a focus and the center, $p$ represents the major radius, and $q$ represents the minor radius of the ellipse. [How did we get this formula?] Finding the focal length Based on the formula, we can see that the ellipse is centered at $(-7,10)$, has a major radius of $\sqrt {13}$, and a minor radius of $\sqrt {8}$. [How did we find this?] Therefore, we can find the focal length, $f$ using the following formula. $\begin{aligned}f^2&=p^2-q^2\\\\ f^2&=(\sqrt{13})^2-(\sqrt{8})^2\\\\ f^2&=5\\\\ f&=\sqrt{5}\end{aligned}$ Therefore, each focus is $\sqrt {5}$ units away from the center of the ellipse. Finding the foci Note that the major radius corresponds to the major axis, which is vertical in this case. So the foci lie $ {\sqrt {5}}$ units to the top and bottom of the center. Therefore, the foci lie at $(-7,10\pm {\sqrt{5}})$, which is to say $(-7,10+\sqrt{5})$ and $(-7,10-\sqrt{5})$. Summary The foci of the ellipse are $(-7,10+\sqrt{5})$ and $(-7,10-\sqrt{5})$.